534 · House Robber II
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Solution31
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Description
After robbing those houses on that street, the thief has found himself a new place for his thievery so that he will not get too much attention. This time, all houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Each house holds a certain amount of money. The only constraint you face is that adjacent houses are equipped with interconnected anti-theft systems, which will automatically alarm when two adjacent houses are robbed on the same day.
Given a list of non-negative integers representing the amount of money of each house, determine the maximum amount of money you can rob tonight without alerting the police.
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This is an extension of House Robber.It just turns a straight line into a circle.
Example
Example1
Input: nums = [3,6,4]
Output: 6
Example2
Input: nums = [2,3,2,3]
Output: 6
解法1：用DP。分house[0]被抢和不被抢两种情况。dp[n]应该还可以空间优化，也就是说只用3个变量prepre, pre, cur就可以。
class Solution {
public:
/**
* @param nums: An array of non-negative integers.
* @return: The maximum amount of money you can rob tonight
*/
int houseRobber2(vector<int> &nums) {
int n = nums.size();
if (n == 0) return 0;
if (n == 1) return nums[0];
int dp[n] = {0}; //dp[i]: the maximum money can be robbed for the first i houses.
//the 0th house is not robbed
dp[1] = nums[1];
for (int i = 2; i < n; i++) {
dp[i] = max(dp[i - 2] + nums[i], dp[i - 1]);
}
int res = dp[n - 1];
//the 0th house is robbed
dp[0] = nums[0];
dp[1] = nums[0]; //注意这一行！
for (int i = 2; i < n - 1; i++) { //0th house is robbed, so i < n - 1
dp[i] = max(dp[i - 2] + nums[i], dp[i - 1]);
}
return max(res, dp[n - 2]);
}
};